. What is the difference between the geometries of the bonding of the two oxygen atoms in acetic acid

Answer:[tex]\rho=7.87g/mL[/tex]

Explanation:

Hello,

In this case, we are analyzing a problem in which the Archimedes principle is considered, thus, as the volume of the metal rod is determined by the difference between the mass of the water and the mass of the water and the rod:

[tex]V_{rod}=15.00mL-10.00mL=5.00mL[/tex]

In such a way, the density turns out:

[tex]\rho =\frac{m_{rod}}{V_{rod}}=\frac{39.35g}{5.00mL} \\ \\\rho=7.87g/mL[/tex]

Regards.

Answer:

2 lone pairs of electrons

Explanation:

In this case, the central atom is oxygen. If we check the periodic table we will have an atomic number of 8. So, the electronic configuration would be:

[tex]1s^2~2s^2~2p^4[/tex]

We have 6 electrons in the last level. Therefore, we have 6 valence electrons. For the hydrogens, we will have only 1 electron each. So, in total, we have 8 electrons.

With this in mind, we have to draw two lone pairs in the oxygen and 1 bond with each hydrogen.

See figure 1.

I hope it helps!

Answer:

7.5 g

Explanation:

There is some info missing. I think this is the original question.

Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

[tex]4.9 g \times \frac{1mol}{98.00g} = 0.050mol[/tex]

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

[tex]0.050mol \times \frac{149.09 g}{mol} = 7.5 g[/tex]

The question is incomplete. Here is the complete question.

Acetic acid and water react to form hydronium cation and acetate anion, like this:

[tex]HCH_{3}CO_{2}_{(aq)}+H_{2}O_{(l)} -> H_{3}O^{+}_{(aq)}+CH_{3}CO_{2}^{-}_{(aq)}[/tex]

Imagine 226 mmol of [tex]CH_{3}CO_{2}^{-}[/tex] are added to a flask containing a mixture of [tex]HCH_{3}CO_{2}[/tex], [tex]H_{2}O[/tex],[tex]H_{3}O^{+}[/tex] and [tex]CH_{3}CO_{2}^{-}[/tex] at equilibrium and then answer the following questions:

1) What's the rate of the reverse reaction before any [tex]CH_{3}CO_{2}^{-}[/tex] has been added to the flask?

a) 0

b) Greater than 0, but less than the rate of the forward reaction

c) Greater than 0, but equal to the rate of the forward reaction

d) Greater than 0, but  greater than the rate of the forward reaction

2) What is the rate of the reverse reaction just after the [tex]CH_{3}CO_{2}^{-}[/tex] has been added to the flask?

a) 0

b) Greater than 0, but less than the rate of the forward reaction

c) Greater than 0, but equal to the rate of the forward reaction

d) Greater than 0, but  greater than the rate of the forward reaction

3) What is the rate of the reverse reaction when the system has again reached equilibrium?

a) 0

b) Greater than 0, but less than the rate of the forward reaction

c) Greater than 0, but equal to the rate of the forward reaction

d) Greater than 0, but  greater than the rate of the forward reaction

4) How much more [tex]CH_{3}CO_{2}^{-}[/tex]  is in the flask when the system has again reached equilibrium?

a) None

b) Some, but less than 226 mmol

c) 226 mmol

d) More than 226 mmol.

Answer: 1) c) Greater than 0, but equal to the rate of the forward reaction

2) d) Greater than 0, but  greater than the rate of the forward reaction

3) c) Greater than 0, but equal to the rate of the forward reaction

4) b) Some, but less than 226 mmol

Explanation: A reversible chemical reaction reaches its equilibrium when forward and reverse reaction are at the same rate. At that point, equilibrium has a constant called K.

Equilibrium constant depends on the concentration of its products and reagents.

For the reaction [tex]HCH_{3}CO_{2}_{(aq)}+H_{2}O_{(l)} -> H_{3}O^{+}_{(aq)}+CH_{3}CO_{2}^{-}_{(aq)}[/tex],

the forward is towards production of [tex]CH_{3}CO_{2}^{-}[/tex] and reverse is towards the production of [tex]HCH_{3}CO_{2}[/tex].

1) When equilibrium is reached, forward and reverse are at the same rate and are different from zero.

2) When adding a compound, the equilibrium is broken. So, to go back to the equilibrium, reaction tend to counteract the change. In the case of acetic acid and water above, when adding acetate anion, the reverse reaction will produce more acetic acid to restore equilibrium, so reverse reaction will be at a greater rate and different from 0.

3) After a while when the system is back to the equilibrium, the rate will be equal again.

4) After second equilibrium, acetate anion will have less mmol than when this new equilibrium state started.

Answer:

In solving this question it is important to note that entropy is referred to as the degree of randomness in a compound or system.

Entropy is highest in gas and lowest in solids

The order is given as:

Gas > Liquid > Solid

Another criteria is the more the elements in a compound, the more the entropy.

For example: KO3 has more entropy than KO2 due to the presence of more elements.

The last criteria is calculating their masses , the one with a greater mass has the higher entropy.

The option with the least entropy will be a solid. Al(s) and NaF(s) are both solids. NaF(s) has more elements in a compound which makes it have a higher entropy than Al(s). This means NaF(s) > Al(s).

The next to consider are the gaseous compounds which are HF(g) , SiF4(g) and SiH4(g). The least entropy will be HF(g) . This is because it has the least number of elements of the 3 compounds.

SiF4(g) and SiH4(g) have equal number of elements in a compound but the one with the highest mass is SiF4(g) because fluorine has a higher mass when compared to hydrogen. This means the compound with the highest entropy is SiF4(g).

The order of increasing entropy is given below:

Al(s) < NaF(s) < HF(g) < SiH4(g) < SiF4(g)

Answer:

[tex]7.5x10^{-3}M[/tex]

Explanation:

Hello,

In this case, since the dissociation barium fluoride is represented at equilibrium by:

[tex]BaF_2(s)\rightleftharpoons Ba^{2+}(aq)+2F^-(aq)[/tex]

Hence, the equilibrium expression is:

[tex]Ksp=[Ba^{2+}][F^-]^2[/tex]

Whereas the molar solubility is represented as the reaction extent [tex]x[/tex]:

[tex]Ksp=[x][2x]^2[/tex]

In such a way, we can solve for [tex]x[/tex]:

[tex]1.7x10^{-6}=4x^3\\\\x=\sqrt[3]{\frac{1.7x10^{-6}}{4} } \\\\x=7.5x10^{-3}M[/tex]

Which as said before, is the molar solubility.

Best regards.

Answer:

the total I think would be 35

Answer:

The answer of the scientific STATEMENT is" Appears That Ants Live In Colonies."

Answer:

A. POBr will be consumed in order to establish a new equilibrium.

Explanation:

POBr3(g) =POBr(g) + Brz(g)

The question is based on Le Chatelier's principle. This principle states that whenever a system at equilibrium is stressed, the system would undergo changes to annul that stress.

If more POBr  is added to the system, it means that the concentration of the products have been increased and as a result the equilibrium has been disturbed. To restore equilibrium, the system would have to shift in a direction that reduces the product concentration.

This direction is the backward direction and in doing so, POBr will be consumed.

The correct option is option A.

The statement that describes the behavior of the system is:

A. POBr will be consumed in order to establish a new equilibrium.

POBr₃(g)  ⇄ POBr(g) + Br₂(g)

According to Le Ch-atelier principle,  if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change to re-establish an equilibrium.

If more POBr is added to the system, it means that the concentration of the products have been increased and as a result the equilibrium has been disturbed. To restore equilibrium, the system would have to shift in a direction that reduces the product concentration.

This direction is the backward direction and in doing so, POBr will be consumed.

Thus, the correct option is option A.

Find more information about Equilibrium here:

brainly.com/question/12270624

Answer:

D

Explanation:

In this time there are made a lot of nuclear bombs so itatters on how what how much damage it makes according to its capacity in my opinion.

Nuclear reaction of which nuclear fusion is a type  which depends on the nucleons that fuse as the total mass of single nucleus is less than that of two original nuclei.

There are two types of nuclear reactions which are nuclear fusion and nuclear fission .They involve the combination and disintegration of the element's nucleus respectively.

In nuclear fission, the nucleus of the atom is bombarded with electrons of low energy which splits the nucleus in to two parts .Large amount of energy is released in the process.It is used in nuclear power reactors as it produces large amount of energy.

In nuclear fusion,on the other hand, is a reaction which occurs when two or more atoms combine to form a heavy nucleus.Large amount of energy is released in the process which is greater than that of the energy which is released in nuclear fission process.

Learn more about nuclear reactions,here:

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Answer:

33.85°C

Explanation:

From the question,

Heat lost by the hotter water = heat gained by the colder water

cm'(t₂-t₃) = cm(t₃-t₁)................. Equation 1

Where c = specific heat capacity of water, m' = mass of hot water, m = mass of cold water, t₁ = Initial temperature of cold water, t₂ = Initial temperature of hot water, t₃ =  final temperature of the mixture.

But since the density of water is constant, and mass varies directly as volume, We can replace the mass of water with the volume of water. i.e,

cv'(t₂-t₃) = cv(t₃-t₁)................. Equation 2

Where v' and v are the volume of hot water and cold water respectively

make t₃ the subject of the equation

t₃ = (v't₂+vt₁)/(v'+v)............ Equation 3

Given: v' = 5.0 L, v = 60 L, t₁ = 30°C, t₂ = 80°C

Substitute these values into equation 3

t₃ = (5×80+60×30)/(60+5)

t₃ = 2200/68

t₃ = 33.85°C

Answer: [tex]T_{f}[/tex] = -29.96 °C

Explanation: A solution has a lower freezing point compared to the pure solvent, due to the solute lowering the vapor pressure of the solvent.

The "new" freezing point is calculated as:

[tex]\Delta T_{f} = K_{f}.m[/tex]

[tex]K_{f}[/tex] is the molal freezing-point depression constant

m is molality concentration, i.e., moles of solute per kilogram of solvent

Before calculating freezing point, let's find moles of ethylene glycol ([tex]C_{2}H_{6}O_{2}[/tex]) in the solution:

Molar mass [tex]C_{2}H_{6}O_{2}[/tex] = 62.08 g/mol

For 500 g:

[tex]n = \frac{500}{62.08}[/tex]

n = 8.05 moles

The molality concentration for 0.5kg of water:

m = [tex]\frac{8.05}{0.5}[/tex]

m = 16.11

The freezing point will be:

[tex]\Delta T_{f} = K_{f}.m[/tex]

[tex]\Delta T_{f} = -1.86*16.11[/tex]

[tex]\Delta T_{f} = -29.96[/tex]

Freezing point of a solution of Ethylene Glycol and Water is -29.96°C

Answer:

2.41 × 10²⁰ atoms

Explanation:

Step 1: Given data

Step 2: Convert the mass of the sample to grams

We will use the relationship 1 g = 1,000 mg.

38.3 mg × (1 g/1,000 mg) = 0.0383 g

Step 3: Calculate the number of atoms in 0.0383 g of molybdenum

0.0383 g × (1 atom/1.59 × 10⁻²² g) =  2.41 × 10²⁰ atom

Answer: Br-CH₂-CH₂-C≡N (3-bromopropanenitrile)

Explanation:

The question tells us to " Deduce the identity of the following compound from the spectral data given. C3H4BrN: H NMR, δ 2.98 (2H, triplet), 3.53 (2H, triplet) 13C NMR, δ 21.05 (triplet), 23.87 (triplet), 118.08 (singlet) (ppm):JR, 2963, 2254 cm" Express your answer as a chemical formula".

Answer: Another Alternative to this is seen below

The Double Bond Equivalent of C₃H₄BrN = 3+1-4/2-1/2+1/2 = 2

2254 cm-1 Vc ≡N [IR analysis]

2963.4 = sp³Vc-H streek [IR analysis]

where the first CH₂ = b = 2.98 (2H,triplet) and

the second CH₂ = a = 3.53 (2H,triplet)

Note:  where both are 1HNMR.

where C = c = 118.08 (singlet)

CH₂ = b = 21.05 (triplet)

CH₂ = a = 23.87 ( triplet)

Note: where all are 13CNMR

Important:

I have attached a copy of the solution to enhance better understanding in case the typed solution isn't clear enough.

Answer: The scientific method attempts to minimize the influence of bias or prejudice in the experimenter. Even the best-intentioned scientists can't escape bias. ... That's the job of the scientific method. It provides an objective, standardized approach to conducting experiments and, in doing so, improves their results.

The correct option is A.

Ordinary sea salt is 97 percent sodium chloride whereas Dead Sea salt is a mixture of lots of different chloride and bromide salts.

Dead Sea Salt differs greatly from other sea salts in mineral content, is made up of sodium chloride with a high percentage of magnesium, sulfates, and potassium. You can bring that ancient, therapeutic experience home with our guaranteed authentic Dead Sea bath salts.

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Answer:

[tex]\text{Rate} = k\, [\mathrm{H_2\, (g)}]\cdot [\mathrm{NO\, (g)}]^2[/tex].

[tex]\rm N_2O\, (g)[/tex] is the intermediate.

Explanation:

Balanced overall reaction: [tex]\rm 2\, H_2\, (g) + 2\, NO\, (g) \to N_2\, (g) + 2\, H_2O\, (g)[/tex].

Proposed mechanism:

Start with the slowest, rate-determining step of the proposed mechanism. Here, the rate-determining step is [tex]\rm H_2\, (g) + 2\, NO\, (g) \to N_2O\, (g) + H_2O \, (g)[/tex].

There are two species on the reactant side of this intermediate reaction: [tex]\rm H_2\, (g)[/tex]  and [tex]\rm NO_2\, (g)[/tex]. The concentration of both of them should be in the rate expression of this step.

On the other hand, the coefficient of [tex]\rm H_2\, (g)[/tex] is one while the coefficient of [tex]\rm NO_2\, (g)[/tex] is two. Therefore, in the rate expression of this step, the concentration of [tex]\rm H_2\, (g)\![/tex] should have a power of one, while the concentration of [tex]\rm NO_2\, (g)\![/tex] should have a power of two.

include the rate constant [tex]k[/tex] to obtain the rate expression of the rate-determining slow step:

[tex]\text{Rate} = k\, [\mathrm{H_2\,(g)}]\cdot [\mathrm{NO\, (g)}]^2[/tex].

Make sure that all species in this rate expression are on the reactant side of the overall balanced reaction. Otherwise, further steps would be required to obtain the rate law of the overall reaction.

Therefore, by this proposed mechanism, the rate law of the overall reaction would be [tex]\text{Rate} = k\, [\mathrm{H_2\,(g)}]\cdot [\mathrm{NO\, (g)}]^2[/tex].

In a proposed reaction mechanism, a species is an intermediate if it appeared in one of the proposed steps, but not in the balanced overall equation.

The two steps of this proposed mechanism mentioned five species:

With the exception of [tex]\rm N_2O\, (g)[/tex], all the other species appeared in the overall balanced equation. Therefore, [tex]\rm N_2O\, (g)\![/tex] is the intermediate.

Answer:

The value is  [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]

Explanation:

From the question we are told that

  The  no of moles of the sample is  n = 0.0447 mole

  The formula weight is  [tex]M = 114 \ g/mol[/tex]

   The  mass of water is  [tex]m = 6.19 *10^{2}\ g[/tex]

   The amount of the fuel is  [tex]f= 6.13*10^{-1} \ nutritional \ Cal[/tex]

   The temperature rise is  [tex]\Delta T = 5.05^o[/tex]

Generally

      [tex]1 \ nutritional \ Cal => 4.184*10^{3} \ kJ/kg[/tex]

=>  [tex]f= 6.13*10^{-1} \ nutritional \ Cal \to x[/tex]

=>     [tex]x = \frac{6.13 *10^{-1} * 4.184 *10^{3}}{1}[/tex]

=>     [tex]x = 2.565 *10^{3} \ kJ/kg[/tex]

Answer:

0 kJ/mol.

Explanation:

Hello,

In this case, since the chemical potential can be represented in terms of the Gibbs free energy of formation:

[tex]\mu = \frac{\Delta _fG}{n}[/tex]

Thus, since the Gibbs free energy of formation of an element is zero, the chemical potential is also zero, or just 0 kJ/mol.

Best regards.

Answer:

The sun warms up parts of the oceans. Warm waters rise just like warm air rises. So, as the warmer ocean waters begin to rise in a particular area, the cooler ocean waters from a different area will move in to replace the warmer ocean waters, and this creates our ocean currents.

Hope it helps!

Answer:

90%

Explanation:

it will be 81/90 ×100/1

81 over 90 × 100 over 1

Answer:

D) Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.

Explanation:

Step 1: Given data

Molarity (M): 2.00 M

Volume (V): 500 mL = 0.500 L

Molar mass of NaCl: 58.44 g/mol

Step 2: Calculate the required moles of NaCl

We will use the following expression.

n = M × V

n = 2.00 mol/L × 0.500 L

n = 1.00 mol

Step 3: Calculate the mass corresponding to 1.00 moles of NaCl

1.00 mol × 58.44 g/mol = 58.44 g

Step 4: Describe the procedure to prepare the solution

Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.

Answer:

D) Measure 58.44g NaCl, dissolve it into 400ml water, and then top off to 500ml in a volumetric flask.

Explanation:

I got it right in class!

Hope  this helps!! :))

Answer:

The correct answer is option C, that is, ΔS and ΔSsurr for the process H2O (s) ⇒ H2O(l) are equal in magnitude and opposite in sign.

Explanation:

The temperature at which solid state of water get transformed into liquid state is termed as the melting point of 0 °C. It can be shown by the reaction:  

H2O (s) ⇒ H2O (l)

The degree of randomness of a molecule is known as entropy. With the transformation of ice into liquid state, there is an increase in randomness. Thus, the value of entropy becomes positive as shown:  

Entropy change (ΔSsys) = ΔSproduct - ΔSreactant

= (69.9 - 47.89) J mol/K

= 22.0 J mol/K

Therefore, the value of entropy change is positive.  

Now the value of entropy for surrounding ΔSsurr will be,  

ΔSsurr = -ΔHfusion/T  

= -6012 j/mol/273

= -22.0 J/molK

Hence, the value of ΔSsurr and ΔSsys exhibit same magnitude with opposite sign.  

Answer:

You would use an electrolytic cell.

Answer:

E. In the external circuit, elections flow from the other compartment to the Cu^2+|Cu^+ compartment

Explanation:

In a galvanic cell, electrons flow from anode to cathode. We must remember that oxidation occurs at the anode and reduction occurs at the cathode. Hence the process; M(s) -------> M^+(aq) + e occurs at the anode.

The electrons lost at the anode are conveyed to the cathode where they are accepted by other chemical species and are reduced to the corresponding reduced species according to the reaction; M^+(aq) + e -----> M(s)

Hence the Cu^2+ | Cu^+ half cell, being the cathode accepts electrons from the other half cell for this reduction reaction to take place, hence the answer

Answer:

HPO₄⁻² +  H₂O  ⇄  PO₄⁻³  + H₃O⁺       Ka3

Explanation:

Phosphoric acid, H3PO4 is a dyprotic acid which undergoes in these three succesive ionization reactions.

The last equillibrium is conditionated to the Ka3

H₃PO₄  +  H₂O  ⇄  H₂PO₄⁻  + H₃O⁺     Ka1

H₂PO₄⁻  +  H₂O  ⇄  HPO₄⁻²  + H₃O⁺   Ka2

HPO₄⁻² +  H₂O  ⇄  PO₄⁻³  + H₃O⁺       Ka3

It is an acid because it release a proton, to make the [H₃O⁺] rise.

pH of solution will always be < 7

H₂PO₄⁻  and  HPO₄⁻²  are amphoteric compounds, which means that they can work as an acid (release protons), or base (take protons)

Answer:

I don't know the second one but one of them is Aristotle

Answer:

Aristrotle and John Dalton

Explanation: